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In this question, the right answer is B. While we get B regardless of whether we consider one root or both roots for the third equation, my question pertains to whether we should consider the second root, that is, $–2$.

$–2$ as a root gives $i\sqrt3$ for both LHS and RHS on substitution without avoiding the negative root, as we do when we are solving radical equations.

A related question: Will be take $\sqrt{-25}$ as $5i$ and ignore $–5i$?

Thank you.

Darsen
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Vikram Johri
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3 Answers3

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Step 1 is to define the function you are analyzing. For each of the equations, you can rewrite the problem as finding $x$ such that $f(x) = 0$ - this puts an additional condition on $x$, which is that it must be in the domain of $f(x)$. This will help weed out any non-solutions

  1. $f(x) = 3x^2 - 27 = 3(x^2-9) = 3(x-3)(x+3)$. The domain of definition is all real numbers, hence the roots of $f(x) = 0$ are $\{\pm 3\}$

  2. $f(x) = (2x-1)^2-(x-1)^2 = x(3x-2)$. The domain of definition is all real numbers, hence the roots of $f(x) = 0$ are $\{\frac{2}{3}, 0\}$

  3. $f(x) = \sqrt{x^2-7}-\sqrt{x-1}$. The domain of definition is $x^2 > 7$, hence the roots of $f(x) = 0$ are $\{3\}$

Dhanvi Sreenivasan
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Lets find the roots for each equation.

  1. The roots are $x=\pm3$
  2. You can expand and re-arrange, or solve for $|2x-1|$=$|x-1|$, that is $2x-1=x-1$ or $2x-1=-(x-1)$. Thus we obtain the roots $x=0$ and $x=\frac{2}{3}$
  3. Here solving $x^2-7=x-1$ or $x^2-x-6=(x-3)(x+2)=0$ gives $x=3$ and $x=-2$. But the domain for $\sqrt{x-1}$ is $x\geq1$, thus we exclude the solution $x=-2$. Note that the domain for $\sqrt{x^2-7}$ is $x\le-\sqrt7\lor x\ge \sqrt7$ and $x=3$ lies in $[\sqrt{7},\infty).$

Thus $B$ is correct.

Alessio K
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  • Thanks, my confusion stems from how to incorporate complex roots when the function is entirely real, that is, made up of real coefficients. Are we saying that we can accept complex solutions that emerge from values within the domain of the function ? – Vikram Johri Oct 07 '20 at 11:47
  • Since the function is entirely real, as well as its domain, then it's solutions must also be real. If we assumed a complex-valued square root, then indeed $x=-2$ would also be a solution. Option $A$ has been eliminated because the last equation has only the root $x=3$ (from their their domains). – Alessio K Oct 07 '20 at 12:31
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How to treat (alleged) complex roots of equations is given by the context you are working in, which seems to be, despite your best efforts to deny it, "equations in the real variable involving real-valued functions". As for the third equation, standard procedure in said context is $$\sqrt{x^2-7}=\sqrt{x-1}\Leftrightarrow\begin{cases} x^2-7\ge0\\ x-1\ge0\\ x^2-7=x-1\end{cases}\Leftrightarrow\begin{cases} x\le-\sqrt7\lor x\ge \sqrt7\\ x\ge1\\ x=-2\lor x=3\end{cases}\Leftrightarrow x=3$$